Problem: Let $h(x)=\dfrac{1}{x^6}$. $h'(1)=$
Answer: The strategy We can first rewrite $h(x)$ as a negative power of $x$. Then, the derivative of $h$ can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is negative.) Once we have $h'(x)$, we can plug $x=1$ into it to find $h'(1)$. Rewriting the fraction as a negative power $h(x)=\dfrac{1}{x^6}=x^{-6}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}h'(x) \\\\ &=\dfrac{d}{dx}\left(x^{-6}\right) \\\\ &=-6x^{-6-1} \gray{\text{The power rule}} \\\\ &=-6x^{-7} \end{aligned}$ Evaluating $h'(x)$ So we found that $h'(x)=-6x^{-7}$, which can also be written as $-\dfrac{6}{x^7}$. Now let's plug ${x=1}$ : $\begin{aligned} -\dfrac{6}{({1})^7}&=-\dfrac{6}{1} \\\\ &=-6 \end{aligned}$ In conclusion, $h'(1)=-6$.